Question

If \[ \tanh x=\frac13, \] then \[ 6\sinh^4x+2\cosh^4x+2\sinh^2x+\cosh^2x= \]

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(4\) \bigskip

Hint:

When \(\tanh x\) is given, let \[ \sinh^2x=k,\qquad \cosh^2x=rk, \] where \(r=\dfrac{\cosh^2x}{\sinh^2x}\). Then use \[ \cosh^2x-\sinh^2x=1 \] to determine both quantities quickly.

Answer 1

The Correct Option is D

Concept: The fundamental hyperbolic identity is \[ \cosh^2x-\sinh^2x=1. \] Also, \[ \tanh x=\frac{\sinh x}{\cosh x}. \] Using the given value of \(\tanh x\), we first find \(\sinh^2x\) and \(\cosh^2x\).

Step 1: Find \(\sinh^2x\) and \(\cosh^2x\). Given \[ \tanh x=\frac13. \] Therefore, \[ \frac{\sinh^2x}{\cosh^2x} = \frac19. \] Let \[ \sinh^2x=k. \] Then \[ \cosh^2x=9k. \] Using \[ \cosh^2x-\sinh^2x=1, \] we get \[ 9k-k=1. \] \[ 8k=1. \] \[ k=\frac18. \] Hence, \[ \sinh^2x=\frac18, \qquad \cosh^2x=\frac98. \]

Step 2: Find \(\sinh^4x\) and \(\cosh^4x\). \[ \sinh^4x = \left(\frac18\right)^2 = \frac1{64}. \] \[ \cosh^4x = \left(\frac98\right)^2 = \frac{81}{64}. \]

Step 3: Substitute into the given expression. Let \[ E= 6\sinh^4x+2\cosh^4x+2\sinh^2x+\cosh^2x. \] Substituting the values, \[ E = 6\left(\frac1{64}\right) + 2\left(\frac{81}{64}\right) + 2\left(\frac18\right) + \frac98. \] \[ = \frac6{64} + \frac{162}{64} + \frac{16}{64} + \frac{72}{64}. \] \[ = \frac{256}{64}. \] \[ =4. \]

Step 4: Write the final answer. \[ \boxed{4} \]